 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-1/15b/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

When

V

is approximately constant,

d V / d ϕ ≃ 0

, and exponential expansion gives the constant Hubble parameter

H = H_{inf}

. Take the spatial Fourier transform

ϕ (x, t) = \int \frac{d ^{3} k}{( 2 π ) ^{3}} ϕ_{k} (t) e^{i k \cdot x} .

Since

\nabla^{2} e^{i k \cdot x} = - k^{2} e^{i k \cdot x},

\ddot{ϕ}_{k} + 3 H_{inf} \dot{ϕ}_{k} + \frac{c ^{2} k ^{2}}{a ^{2}} ϕ_{k} = 0 .

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